Capacitors

JEE Main level, single correct, medium

142 of 180

Question 142 of 180

Let CC be the capacitance of a capacitor discharging through a resistor R. Suppose t1t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2t_1 / t_2 will be

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Answer: (C) 14\frac{1}{4}

U=12q2C=12C(q0e−t/T)2=q022Ce−2t/T(\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{1}{2 \mathrm{C}}\left(\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}\right)^2=\frac{\mathrm{q}_0^2}{2 \mathrm{C}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{T}} \quad( where τ=CR)\tau=\mathrm{CR})
U=Uie−2t/τ\mathrm{U}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{\tau}}
12Ui=Uie−2t1/τ\frac{1}{2} \mathrm{U}_{\mathrm{i}}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t}_1 / \mathrm{\tau}}
12=e−2t1/τ⇒t1=T2ln⁡2\frac{1}{2}=\mathrm{e}^{-2 \mathrm{t}_1 / \tau} \Rightarrow \mathrm{t}_1=\frac{\mathrm{T}}{2} \ln 2
Now q=q0e−t/T\quad \mathrm{q}=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}
14q0=q0e−t/2 T\frac{1}{4} \mathrm{q}_0=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{~T}}
t2=T⁡ln⁡4=2T⁡ln⁡2\mathrm{t}_2=\operatorname{T} \ln 4=2 \operatorname{T} \ln 2
∴t1t2=14\therefore \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{1}{4}

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